Question

In the following figures, ABCD is a parallelogram. A circle through A, B, C intersects CD or CD produced at E. Prove that AE = AD.

​

Answer 1

Given: ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E.

To prove: AE = AD

Proof:

∠AED + ∠ABC = 180° ... (1) (Sum of opposite angles of cyclic quadrilaterals is 180°)

∠ADE + ∠ADC = 180° ... (2) (linear pair)

∠ABC = ∠ADC ... (3) (opposite angles of parallelogram are equal)

From (1) and (2)

∠AED + ∠ABC = ∠ADE + ∠ADC

⇒∠AED = ∠ADE (using (3))

In ∆AED,

∠AED = ∠ADE

⇒AD = AE (equal sides have equal angles opposite to them)

Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Given: ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E.

To prove: AE = AD

Proof:

∠AED + ∠ABC = 180° ... (1) (Sum of opposite angles of cyclic quadrilaterals is 180°)

∠ADE + ∠ADC = 180° ... (2) (linear pair)

∠ABC = ∠ADC ... (3) (opposite angles of parallelogram are equal)

From (1) and (2)

∠AED + ∠ABC = ∠ADE + ∠ADC

⇒∠AED = ∠ADE (using (3))

In ∆AED,

∠AED = ∠ADE

⇒AD = AE (equal sides have equal angles opposite to them)